Question

Find the zeros of the following quadratic polynomial and verify the relationship between the zeros and the coefficients:

`q(y) = 7y^2 - 11/3y - 2/3`

Answer 1

Given: `q(y) = 7y^2 - 11/3y - 2/3`.

Step-wise calculation:

1. Clear fractions: multiply equation q(y) = 0 by 3:

21y² – 11y – 2 = 0

2. Use the quadratic formula

`y = (-(-11) ± sqrt((-11)^2 - 4 xx 21 xx (-2)))/(2 xx 21)` 

= `(11 ± sqrt(121 + 168))/42` 

= `(11 ± sqrt(289))/42` 

= `(11 ± 17)/42`

3. So the two zeros are `y_1 = (11 + 17)/42` 

= `28/42`

= `2/3`

`y_2 = (11 - 17)/42` 

= `-6/42`

= `-1/7`

Zeros: `y = 2/3` and `y = -1/7`.

Verify relations:

Sum: `y_1 + y_2 = 2/3 + (-1/7)`

= `(14 - 3)/21` 

= `11/21` 

Also `(-"Coefficient of"  y)/("Coefficient of"  y^2) = -(-11/3)/7`

= `(11/3)/7`

= `11/21`

Product: `y_1 xx y_2 = (2/3)(-1/7)`

= `-2/21` 

Also `("Constant term")/("Coefficient of"  y^2)`

= `(-2/3)/7`

= `-2/21`

Therefore the zeros are `2/3` and `-1/7` and the sum/product relations with the coefficients are verified.